Although Hatcher's result is correct, his "cancellation" explanation needs clarification.
For a singular $m$-simplex $\tau : \Delta^m \to X$, where $\Delta^m = [u_0,\dots,u_m]$, we write $d_k(\tau) = \tau \mid [u_0,\dots,\hat{u}_k,\ldots,u_m]$ for $k = 0,\ldots, m$.
Let us abbreviate $F \circ (\sigma \times \mathbb I)$ by $\phi$. Then we have
$$P(\sigma) = \sum_{i=0}^n (-1)^i \phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]$$
$$\partial P(\sigma) = \sum_{i=0}^n (-1)^i \partial (\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) \\= \sum_{i=0}^n (-1)^i \sum_{j=0}^{n+1} (-1)^jd_j(\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) \\= \sum_{i,j} (-1)^{i+j}d_j(\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) \\ = \sum_{j \le i \le n} (-1)^{i+j}d_j(\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) + \\ \sum_{n+1 \ge j > i} (-1)^{i+j}d_j(\phi \mid [v_0,\ldots,v_i,w_i,\ldots,w_n]) \\= \sum_{j \le i \le n} (-1)^{i+j} \phi \mid [v_0,\ldots,\hat{v}_j,\ldots,v_i,w_i,\ldots,w_n] + \\ \sum_{n+1 \ge j > i} (-1)^{i+j}\phi \mid [v_0,\ldots,v_i,w_i,\ldots,\hat{w}_{j-1},\ldots,w_n] \\ =
\sum_{j \le i \le n} (-1)^{i+j} \phi \mid [v_0,\ldots,\hat{v}_j,\ldots,v_i,w_i,\ldots,w_n] + \\ \sum_{n \ge j \ge i} (-1)^{i+j+1}\phi \mid [v_0,\ldots,v_i,w_i,\ldots,\hat{w}_{j},\ldots,w_n]$$
Now you see that each term indexed by $(i,i)$ with $n \ge i > 0$ in the first sum cancels with the term indexed by $(i-1,i-1)$ in the second sum. Thus only the terms indexed by $(0,0)$ in the first sum and by $(n,n)$ in the second sum survive. These are $\phi \mid [w_0,\ldots,w_n] = g_\#(\sigma)$ and $-\phi \mid [v_0,\ldots,v_n] = -f_\#(\sigma)$. Thus we get
$$\partial P(\sigma) - g_\#(\sigma) + f_\#(\sigma) = \\\sum_{j < i \le n} (-1)^{i+j} \phi \mid [v_0,\ldots,\hat{v}_j,\ldots,v_i,w_i,\ldots,w_n] +\\ \sum_{n \ge j > i} (-1)^{i+j+1}\phi \mid [v_0,\ldots,v_i,w_i,\ldots,\hat{w}_{j},\ldots,w_n]$$
For each face $d_j(\sigma) = \sigma \mid [v_0,\ldots,\hat v_j,\ldots, v_n]$ we have to consider the prism with bottom $[v_0,\ldots,\hat v_j,\ldots, v_n]$ and top $[w_0,\ldots,\hat w_j,\ldots, w_n]$. We get
$$P \partial \sigma = P(\sum_{j=0}^n (-1)^j d_j(\sigma)) = \sum_{j=0}^n (-1)^j P(d_j(\sigma)) = \sum_{j=0}^n (-1)^j P(\sigma \mid [v_0,\ldots,\hat v_j,\ldots, v_n])=\\ \sum_j (-1)^j \left(\sum_{i<j}(-1)^i \phi \mid [v_0,\ldots,v_i,w_i,\ldots\hat w_j,\ldots, w_n] + \\ \sum_{i\ge j}(-1)^i \phi \mid [v_0,\ldots,\hat v_j,\ldots, v_{i+1},w_{i+1},\ldots w_n]\right) \\= \sum_{i < j} (-1)^{i+j} \phi \mid [v_0,\ldots,v_i,w_i,\ldots\hat w_j,\ldots, w_n] + \\ \sum_{i > j} (-1)^{i+j+1} \phi \mid [v_0,\ldots,\hat v_j,\ldots, v_{i},w_{i},\ldots w_n]$$
Best Answer
I understand now what the wording is trying to convey; it escaped me at first. The message is that pairs $(i,j)$ in the first sum with $i=j$ match up and cancel with pairs $(i-1,j-1)$ in the second sum; that is, the pair $(i=2,j=2)$ in the first sum over $j\leq i$ matches up with the pair $(i=1,j=1)$ in the second sum over $i\leq j$. I will leave this up in case anyone else runs into similar confusion.